# Parallel transporting of a vector in curved space

Am I correct in saying that the angular deficit (change in angle) of a vector transported around a closed surface on a curved surface can only be observed by flattening the surface?

Actually a further problem- I understand it from the flat sheet to a cone: Cut out a pie from a sheet, Draw a (incomplete) circle around the tip of the pie, Draw arrows pointing in the same direction (parallel) along the circle (i.e. parallel transport on the flat surface with the pie cut out) and now roll the sheet into a cone with the pie tip as cone tip and matching the edges of the pie. The arrows are no longer parallel.

But I have not understood it starting from a cone. Now the surface is curved, what is a parallel vector, does it stay on the surface? How do I try to parallel transport a vector on the circle on a cone? Any clear examples available.

I sort of do understand the Tensor math. I am looking for a visual explanation.

## Answers and Replies

tiny-tim
Science Advisor
Homework Helper
hi say_cheese! Am I correct in saying that the angular deficit (change in angle) of a vector transported around a closed surface on a curved surface can only be observed by flattening the surface?

you can't flatten a surface!! :yuck:

(except for very special surfaces, such as a cone)

and if you can, then the angular deficit will always be zero Dale
Mentor
2020 Award
Am I correct in saying that the angular deficit (change in angle) of a vector transported around a closed surface on a curved surface can only be observed by flattening the surface?
No. Consider a vector parallel transported along a sphere. Suppose it starts at the equator on the prime meridian pointing north. It is transported 90 degrees west, then north to the pole, then south to the starting point. At the end it is pointing east.

But I have not understood it starting from a cone. Now the surface is curved, what is a parallel vector, does it stay on the surface?
Yes, it always stays on the surface.

I guess I can flatten the surface (as in Lambert Conformal conic ??) with a missing piece just like in a cone. The edge of the missing pie will not be straight, but vary sinusoidally (90 to latitude of the circle drawn). Am I right?

No. Consider a vector parallel transported along a sphere. Suppose it starts at the equator on the prime meridian pointing north. It is transported 90 degrees west, then north to the pole, then south to the starting point. At the end it is pointing east.

Yes, it always stays on the surface.

Great! I get it.

tiny-tim
Science Advisor
Homework Helper
I guess I can flatten the surface (as in Lambert Conformal conic ??)

no, the lambert conformal conic does not flatten the surface, any more than mercator does …

you can't flatten a sphere!!

pervect
Staff Emeritus
Science Advisor
I should also mention in passing that a cone doesn't have any intrinsic curvature, which is the sort that's of interest in GR. (The non-technical term "curvature" is a bit ambiguous - but in the parallel tranpsort sense we're talking about, you need intrinsic curvature to see any effect on parallel tranpsorted vectors).

A cone is basically a flat piece of - paper, or whatever - pasted together in a certain way. So it's n ot intrinsically curved, unlike a sphere.

IF you try to flatten a sphere made out of non-stretchy paper, you'll see that the paper tears - another sign that the sphere is curved, and the cone isn't.

I think parallel transport is used in general relativity, it defines the Riemann tensor. However in GR this implies the use of a closed curve in space-time which I do not understand it seems paradoxical.

HallsofIvy
Science Advisor
Homework Helper
ut I have not understood it starting from a cone. Now the surface is curved, what is a parallel vector, does it stay on the surface?

Yes, it always stays on the surface.
More correctly it stays on the tangent plane to the surface.

HallsofIvy
Science Advisor
Homework Helper
I guess I can flatten the surface (as in Lambert Conformal conic ??) with a missing piece just like in a cone. The edge of the missing pie will not be straight, but vary sinusoidally (90 to latitude of the circle drawn). Am I right?
You can flatten only "rectifiable surfaces" which are surfaces that, like a plane, cylinder, or cone, have the property that, through each point on the surface, there exist at least one line that lies completely in the surface.

Dale
Mentor
2020 Award
I think parallel transport is used in general relativity, it defines the Riemann tensor. However in GR this implies the use of a closed curve in space-time which I do not understand it seems paradoxical.
Parallel transport is more general than parallel transport around a closed curve. I.e. parallel transport is also key to the definition of the covariant derivative, which is defined along open curves. It is primarily in the definition of curvature where parallel transport around closed curves is used.

As you indicate, it is not possible to parallel transport a timelike object, like a gyroscope, along a closed path in spacetime, nor around a spacelike open path. However, since parallel transport is also defined along open paths the effects of curvature can be measured along open paths as well.

flattening has to do with orientability and nonorientability of surfaces.parallel transport definition of levi civita takes care both of them.

Thanks for the clarifications. The key thing seems to be that in order to measure the curvature, one has to transport in two or more directions. (Orienting a vector to the north at Arctic circle and transporting it around a longitude circle does not rotate the vector, except for great circles like longitudes and equator).

no, the lambert conformal conic does not flatten the surface, any more than mercator does …

you can't flatten a sphere!!

True! I do see that slicing the sphere in one dimension opens up in that dimension, but the other remains curved, like peeled orange skin remains spherical.

No. Consider a vector parallel transported along a sphere. Suppose it starts at the equator on the prime meridian pointing north. It is transported 90 degrees west, then north to the pole, then south to the starting point. At the end it is pointing east.

Yes, it always stays on the surface.

I got the case of travelling to the pole and returning to the starting point, that is a triangular path, but does this explanation work when working it out on a trapezoidal path, say west along equator then north to 30 deg latitude and then return? Would you not keep pointing the vector in the north direction when returning to the starting longitude at the latitude of 30 deg., in which case, there is no rotation of the vector?

Dale
Mentor
2020 Award
I got the case of travelling to the pole and returning to the starting point, that is a triangular path, but does this explanation work when working it out on a trapezoidal path, say west along equator then north to 30 deg latitude and then return? Would you not keep pointing the vector in the north direction when returning to the starting longitude at the latitude of 30 deg., in which case, there is no rotation of the vector?
There are special cases where you get no rotation, for instance transport along a great circle will bring it back without rotation. However, the fact that there is ANY loop which does not bring it back shows that the manifold is curved.

However, your counter-example actually doesn't work. If you parallel transport a north pointing vector along any latitude line other than the equator (which is a great circle) then it will stop pointing north. Consider parallel transporting around the 89° lattitude line. When you start, the vector is pointing straight at the north pole, but if you parallel transport it 90° W then it will be approximately pointing W. It isn't as obvious, but the same thing happens along the 30° latitude line.

Remember that parallel transport preserves dot products and that geodesics parallel transport their tangent vector. So a vector will have the same dot product with the tangent vector when transported along a geodesic, but not along other paths. So, if you transport a north vector eastward along the equator (which is a geodesic) then it starts with a 0 dot product with the tangent and ends with a 0 dot product with the tangent. Since the tangent is east in both cases then the vector is north in both cases. But if you transport a north vector eastward along another latitude line (which is not a geodesic) then it starts with a 0 dot product with the tangent and ends with a non-0 dot product with the tangent. Since the tangent is east in both cases then the vector is not north in the end.

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Remember that parallel transport preserves dot products and that geodesics parallel transport their tangent vector. So a vector will have the same dot product with the tangent vector when transported along a geodesic, but not along other paths. So, if you transport a north vector eastward along the equator (which is a geodesic) then it starts with a 0 dot product with the tangent and ends with a 0 dot product with the tangent. Since the tangent is east in both cases then the vector is north in both cases. But if you transport a north vector eastward along another latitude line (which is not a geodesic) then it starts with a 0 dot product with the tangent and ends with a non-0 dot product with the tangent. Since the tangent is east in both cases then the vector is not north in the end.

Thanks. I was actually going from the other way about the dot product and the geodesic. I understood that the dot product remains constant (also zero) along a geodesic. Now, I see it. At other latitudes the north vector which is tangential to the constant latitude line is not perpendicular to the east- west direction, i.e .constant longitude line This is, I guess, the reason why one can invoke a tangent cone to explain the rotation of the vector.

pervect
Staff Emeritus
Science Advisor
I'll mention a little known technique for graphically parallel transporting vectors, known as Schild's Ladder.

You'll find the original textbook source in MTW's book "Gravitation".

To apply it you need to know how to draw geodesic curves on your surface, and be able to measure out equal intervals along the curves. If we can measure distances, "equal intervals" just means "equal distances".

You can think of geodesic curves as being analogous to "straight lines" for this exercise, due to the property of geodesics that they are the curve of the shortest distance lying wholy on the surface that connects two points.

Because geodesic curves on a sphere are great circles, one has all the tools one needs to use this technique to parallel transport curves along a sphere.

We start with some general curve AB, which does not have to be a geodesic, and some vector, AC. We want to parallel transport AC along the curve to B. But we need to do this in steps.

Only one of many such needed steps is shown on the drawing.

The first step consists of finding some point A' on the curve AB that's close to A. Then the curve segment AA' will be basically "straight", i.e. a geodesic.

Next we find the midpoint of A'C, and label that d.

We draw a geodesic ("straigh line") from A to d and beyond. We mark the point C' such that Ad == dC' as per the diagram, i.e. point d is the midpoint of AC'

Then, in the limit as we make A' close to A, A'C' is the parallel transported version of AC. We have to repeat the step many times in order to get all the way to B, however.

This works because in the absence of torsion (something that GR presupposes), a 4-sided geometric figure (i.e. a quadrilateral) with two pairs of equal-length sides is a parallelogram. And the sides of a parallelogram are parallel.

This technique is rather similar to a graphical device known as a pantograph

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